POJ2251 Dungeon Master —

POJ2251 Dungeon Master —— BFS

阅读量：4926 次

发布时间：2019-06-11

本文共 3522 字，大约阅读时间需要 11 分钟。

题目链接：

Dungeon Master

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 37296		Accepted: 14266

Description

You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or down. You cannot move diagonally and the maze is surrounded by solid rock on all sides.

Is an escape possible? If yes, how long will it take?

Input

The input consists of a number of dungeons. Each dungeon description starts with a line containing three integers L, R and C (all limited to 30 in size).

L is the number of levels making up the dungeon.

R and C are the number of rows and columns making up the plan of each level.

Then there will follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock is indicated by a '#' and empty cells are represented by a '.'. Your starting position is indicated by 'S' and the exit by the letter 'E'. There's a single blank line after each level. Input is terminated by three zeroes for L, R and C.

Output

Each maze generates one line of output. If it is possible to reach the exit, print a line of the form

Escaped in x minute(s).

where x is replaced by the shortest time it takes to escape.

If it is not possible to escape, print the line

Trapped!

Sample Input

3 4 5S.....###..##..###.#############.####...###########.#######E1 3 3S###E####0 0 0

Sample Output

Escaped in 11 minute(s).Trapped!

Source

题解：

三维的纯BFS。

代码如下：

1 #include 
      
          2 #include 
       
           3 #include 
        
            4 #include 
         
             5 #include 
          
              6 #include 
           
             7 #include 
            
              8 #include 
             
               9 #include 
               10 #include 
               
                 11 #include 
                
                  12 #define ms(a,b) memset((a),(b),sizeof((a))) 13 using namespace std; 14 typedef long long LL; 15 const int INF = 2e9; 16 const LL LNF = 9e18; 17 const int MOD = 1e9+7; 18 const int MAXN = 30+10; 19 20 struct node 21 { 22 int x, y, z, step; 23 }; 24 25 int L, R, C; 26 char m[MAXN][MAXN][MAXN]; 27 int vis[MAXN][MAXN][MAXN]; 28 int dir[6][3] = { { 1,0,0},{ 0,1,0},{ 0,0,1},{-1,0,0},{ 0,-1,0},{ 0,0,-1} }; 29 30 queue
                 
                  que; 31 int bfs(node s, node e) 32 { 33 ms(vis,0); 34 while(!que.empty()) que.pop(); 35 36 s.step = 0; 37 vis[s.x][s.y][s.z] = 1; 38 que.push(s); 39 40 while(!que.empty()) 41 { 42 node now = que.front(); 43 que.pop(); 44 45 if(now.x==e.x && now.y==e.y && now.z==e.z) 46 return now.step; 47 48 for(int i = 0; i<6; i++) 49 { 50 node tmp; 51 tmp.x = now.x + dir[i][0]; 52 tmp.y = now.y + dir[i][1]; 53 tmp.z = now.z + dir[i][2]; 54 if(tmp.x>=1 && tmp.x<=L && tmp.y>=1 && tmp.y<=R && tmp.z>=1 && tmp.z<=C 55 && m[tmp.x][tmp.y][tmp.z]!='#' && !vis[tmp.x][tmp.y][tmp.z] ) 56 { 57 vis[tmp.x][tmp.y][tmp.z] = 1; 58 tmp.step = now.step + 1; 59 que.push(tmp); 60 } 61 } 62 } 63 return -1; 64 } 65 66 int main() 67 { 68 while(scanf("%d%d%d",&L, &R, &C) && (L||R||C)) 69 { 70 for(int i = 1; i<=L; i++) 71 for(int j = 1; j<=R; j++) 72 scanf("%s", m[i][j]+1); 73 74 node s, e; 75 for(int i = 1; i<=L; i++) 76 for(int j = 1; j<=R; j++) 77 for(int k = 1; k<=C; k++) 78 { 79 if(m[i][j][k]=='S') s.x = i, s.y = j, s.z = k; 80 if(m[i][j][k]=='E') e.x = i, e.y = j, e.z = k; 81 } 82 83 int ans = bfs(s,e); 84 if(ans==-1) 85 printf("Trapped!\n"); 86 else 87 printf("Escaped in %d minute(s).\n", ans); 88 } 89 return 0; 90 }

View Code

转载于:https://www.cnblogs.com/DOLFAMINGO/p/7538599.html

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